tìm x , biết:
(x-2)2 - 4x + 8 =0
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29 tháng 11 2023
a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
TQ
1
T
11 tháng 12 2023
a) $(x-3)^2-(x+2)(x-2)=-5$
$\Rightarrow x^2-2\cdot x\cdot3+3^2-(x^2-2^2)=-5$
$\Rightarrow x^2-6x+9-(x^2-4)=-5$
$\Rightarrow x^2-6x+9-x^2+4=-5$
$\Rightarrow-6x+13=-5$
$\Rightarrow-6x=-18$
$\Rightarrow x=3$
b) $x^3-2x^2-4x+8=0$
$\Rightarrow(x^3-2x^2)-(4x-8)=0$
$\Rightarrow x^2(x-2)-4(x-2)=0$
$\Rightarrow (x^2-4)(x-2)=0$
$\Rightarrow (x^2-2^2)(x-2)=0$
$\Rightarrow (x-2)(x+2)(x-2)=0$
$\Rightarrow (x-2)^2(x+2)=0$
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
$\text{#}Toru$
TN
5 tháng 7 2016
a)8x2+30x+7=0
=>8x2+28x+2x+7=0
=>(8x2+2x)+(28x+7)=0
=>2x(4x+1)+7(4x+1)=0
=>(2x+7)(4x+1)=0
\(\Rightarrow\orbr{\begin{cases}x=-\frac{7}{2}\\x=-\frac{1}{4}\end{cases}}\)
b)(x2-4x)2-8(x2-4x)+15=0
=>x4-8x3+8x2+32x+15=0
=>(x-5)(x+1)(x2-4x-3)=0
\(\Rightarrow\hept{\begin{cases}x=5\\x=-1\\x=2-\sqrt{7};x=\sqrt{7}+2\end{cases}}\)
HT
0
Y
21 tháng 7 2023
\(a,3\sqrt{x}-7=0\left(dk:x\ge0\right)\\ \Leftrightarrow3\sqrt{x}=7\\ \Leftrightarrow\sqrt{x}=\dfrac{7}{3}\\ \Leftrightarrow x=\dfrac{49}{9}\left(tmdk\right)\)
Vậy \(S=\left\{\dfrac{49}{9}\right\}\)
\(b,\sqrt{x-2}+\sqrt{4x-8}=3\left(dk:x\ge2\right)\\ \Leftrightarrow\sqrt{x-2}+\sqrt{4\left(x-2\right)}=3\\ \Leftrightarrow\sqrt{x-2}+2\sqrt{x-2}=3\\ \Leftrightarrow3\sqrt{x-2}=3\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\\ \Leftrightarrow x=3\left(tmdk\right)\)
Vậy \(S=\left\{3\right\}\)
21 tháng 7 2023
a: =>3*căn x=7
=>căn x=7/3
=>x=49/9
b: =>3*căn x-2=3
=>căn x-2=1
=>x-2=1
=>x=3
DN
0
NT
3
23 tháng 8 2021
3) \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
(x-2)^2-4x+8=0
=>(x-2)^2-4(x-2)=0
=>(x-2)(x-2-4)=0
=>(x-2)(x-6)=0
=>x=2 hoặc x=6
`(x-2)^2 -4x+8=0`
`<=> (x-2)^2 -(4x-8)=0`
`<=> (x-2)^2 - 4(x-2)=0`
`<=> (x-2)(x-2-4)=0`
`<=>(x-2)(x-6)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)